How Do I Correctly Calculate the Speed of Casing in a Planetary Gear System?

[andycampbell1]

Homework Statement

Hi

I am doing this gear problem, I have attached the question and how I've started this off I am having problems working it all out as the question said I should try and have the casing rotating at a speed of 15-25 rpm.

I have, these gear sizes which were obtained from a manufacturers catalogue.

TB = 36
TC = 24
TD = 40
TE = 80
NX = 100 rpm
NY = 350 rpm

The Attempt at a Solution

To work out the number of teeth on casing F, from the geometry of the figure
dF/2 = dE/2+dB
so dF= dE + 2dB
dF= 80+ 2(40) = 160 teeth.

Using my my table of motion which I have attached for the speed of gear F, there is a mistake on it as the 2 TC's cancel out to end up y-x*(TB/TF) I have used the input speed for x and the output speed for y so my equation works out as 350-100 * (36/160) = 327.5 rpm this is obviously very wrong. Where have I went wrong? Have I started at the wrong point?

Thanks in advance.

 

[Valkarie]

Homework EquationsdF/2 = dE/2+dBx-y*(TB/TF)The Attempt at a SolutionTo calculate the speed of gear F, you need to use the gear ratio equation:x-y*(TB/TF)Where x is the input speed (NX), y is the output speed (NY), TB is the number of teeth on gear B (36), and TF is the number of teeth on gear F (160).Plugging in the values, we get:100 - 350*(36/160) = 25 rpmThis means that the speed of gear F will be 25 rpm, which is within the desired range of 15-25 rpm.